What is the derivative of f(t) = (e^(1-t) , tan^2t ) ?

1 Answer
Nov 21, 2017

See explanation.

Explanation:

It depends on what derivative you want. If you want dy/(dt) and (dx)/dt, simply differentiate the x and y functions with respect to t, the same way you would differentiate a function f(x) with respect to x.

x(t)= e^(1-t) -> dx/dt = -e^(1-t)
y(t) = tan^2(t) = (tan t)(tan t) -> dy/dt = 2tan(t)d/dt(tan t) = (2 tan t)/(sec^2(t)

If you specifically want dy/dx, we must use the chain rule .

dy/dx = dy/dt dt/dx

This last term is simply one over dx/dt...

dy/dx = (2tant)/(sec^2(t)) * 1/(-e^(1-t)

Recalling that e^(a-b) = e^a/e^b...

dy/dx = (2tant)/(sec^2(t)) * -1/((e/e^t)) = (2tant)/(sec^2(t)) * -e^t/e = (2tant)/(sec^2(t)) * -e^(t-1) = ((-2tant)(e^(t-1)))/(sec^2(t))